Answer to Question #121105 in Differential Equations for Joseph Se

Question

1. Solve the differential equation. Use the fact that the given equation is homogeneous.
dy / dx = (x^2 + 8y^2) / 3xy

2. A certain college graduate borrows 7486 dollars to buy a car. The lender charges interest at an annual rate of 19% . Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 2 years. Also determine how much interest is paid during the 2-year period.

Round your answers to two decimal places.

A. Payment rate ______ dollars per year. (absolute tolerance +/- 0.01)
B. Interest Paid _____ dollars (absolute tolerance is +/- 0.05)

Answer 1

1) $\frac {dy}{dx}$ = $\frac {x²+8y²}{3xy}$ = $\frac {1+8(y/x)²}{3y/x}$

Since $\frac {dy}{dx}$ is a function of y/x , it is homogeneous form.

To solve this

Put y = vx

Differentiating $\frac {dy}{dx}$ = v + x$\frac {dv}{dx}$

So v + $x\frac {dv}{dx}$ = $\frac {1+8(v)²}{3v}$

=> $x\frac {dv}{dx}$ = $\frac {1+8(v)²}{3v}$ - v = $\frac {1+5v²}{3v}$

=> $\frac {vdv}{5v²+1} = \frac {dx}{3x}$

=> $\frac {vdv}{v²+\frac {1}{5}} = \frac {5}{3}\frac {dx}{x}$

Integrating,

$\int\frac {vdv}{v²+\frac {1}{5}} =\int \frac {5}{3}\frac {dx}{x}$

Let v²+$\frac {1}{5}$ = z

So 2 v dv = dz

$\frac {1}{2}\int\frac {dz}{z} =\ \frac {5}{3}\int\frac {dx}{x}$

$3\int\frac {dz}{z} =10\int\frac {dx}{x}$

=> 3 ln |z |= 10 ln | x| + lnA

=> ln|z|³= ln x¹⁰+lnA

=> ln |z|³ = ln (Ax¹⁰)

=> |z|³ = Ax¹⁰

=> |v²+ $\frac {1}{5}$ |³ = Ax¹⁰

=> | 5v²+1|³ = 125Ax¹⁰

=> | 5y²+ x²|³ = 125Ax⁶x¹⁰

=> (5y²+x²)³ = C³x¹⁶ Considering 125A = C³

=> (x²+5y²) = C$x^{\frac {16}{3}}$

This is the general solution of given differential equation

2) Let x be the amount borrowed by the student and t represents time

So $\frac {dx}{dt}$ = $\frac {19x}{100} - k$ = 0.19x - k

=> $\frac {dx}{0.19x-k} = dt$

Integrating,

$\int\frac {dx}{0.19x-k} = \int dt$

$\frac {1}{0.19}$ ln |0.19x-k| = t + C

As the loan is repayable , 0.19x < k

So ln (k-0.19x) = 0.19t + 0.19C

When t = 0, x = 7486

So 0.19C = ln(k - 0.19*7486)

=> ln (k-0.19x) = 0.19t + ln(k - 0.19*7486)

As the borrower wants to pay off the loan in 2 years, when t = 2 , x = 0

So ln k = 0.38 + ln(k - 0.19*7486)

=> ln $\frac { k} { k - 0.19*7486}$ = 0.38

=> k = (k - 0.19*7486)e^0.38

=> k(e^0.38-1) = 0.19*7486*e^0.38

=> k = $\frac {0.19*7486*e^{0.38}}{e^{0.38}-1}$

=> k = 4499.10

As loan is repaying at a constant rate of 4499.10 dollars per year, inv2-year period borrower pays 2*4499.10 dollars = 8998.20 dollars.

So interest in 2-year period

= Amount paid in 2 years subtracted by amount of loan taken

= 8998.20 - 7486 dollars

= 1512.20 dollars

Answer

A. Payment rate 4499.10 dollars per

year

B. Interest paid 1512.20 dollars